Calculus II 
Test #1

March 5, 2002

Name____________________ 
R. Hammack

Score ______

(2) A object moving on a straight line has a (nonconstant) acceleration
of 6t feet per second per second at time t.
It travels a distance of 33 feet between times t = 0 and t = 3. Find
the object's velocity function v(t).
a(t) = 6t
v(t) =
= 3+C (We
just need to find C)
s(t) =
= +Ct
+K
From s(0) = 0 we get K = 0, so s(t)
= +Ct
Now, 33 = s(3) =
gives the equation 33 = 27 + 3C, or 3C = 6, so C = 2.
Thus v(t) = 3+2
(3)
(4) Find the average value of the function on
the interval [0, ln(2)].
Using the formula for average value, the average value is
=
==
(5) 20
Over the interval [6,2], the graph of y = x forms two triangles. Adding their
areas, we get the answer of
(1/2)(2)(2) + (1/2)(6)(6) = 2 + 18 = 20.
(6) The graph of a function f(x) has slope at
each point (x, f(x)).
Also, the graph of f(x) passes through the point (1,
5). Find the function f(x).
The information says slope = f '(x) = .
This means f(x) =
.
Thus f(x) =
To determine f completely, we just need to find C. To do this,
use the fact that f(1)=5.
5=f(1)=
5= 0 + 3/4 +C
Therefore C = 17/4, and the answer to the question is
f(x) =
(7) The expression represents
a definite integral over the interval [3, 5]. Write the definite
integral. (You do not need to find its value.) ANSWER: